--- /dev/null
+DCPU-16 Specification\r
+Copyright 2012 Mojang\r
+Version 1.1 (Check 0x10c.com for updated versions)\r
+\r
+* 16 bit unsigned words\r
+* 0x10000 words of ram\r
+* 8 registers (A, B, C, X, Y, Z, I, J)\r
+* program counter (PC)\r
+* stack pointer (SP)\r
+* overflow (O)\r
+\r
+In this document, anything within [brackets] is shorthand for "the value of the RAM at the location of the value inside the brackets".\r
+For example, SP means stack pointer, but [SP] means the value of the RAM at the location the stack pointer is pointing at.\r
+\r
+Whenever the CPU needs to read a word, it reads [PC], then increases PC by one. Shorthand for this is [PC++].\r
+In some cases, the CPU will modify a value before reading it, in this case the shorthand is [++PC].\r
+\r
+Instructions are 1-3 words long and are fully defined by the first word.\r
+In a basic instruction, the lower four bits of the first word of the instruction are the opcode,\r
+and the remaining twelve bits are split into two six bit values, called a and b.\r
+a is always handled by the processor before b, and is the lower six bits.\r
+In bits (with the least significant being last), a basic instruction has the format: bbbbbbaaaaaaoooo\r
+\r
+\r
+\r
+Values: (6 bits)\r
+ 0x00-0x07: register (A, B, C, X, Y, Z, I or J, in that order)\r
+ 0x08-0x0f: [register]\r
+ 0x10-0x17: [next word + register]\r
+ 0x18: POP / [SP++]\r
+ 0x19: PEEK / [SP]\r
+ 0x1a: PUSH / [--SP]\r
+ 0x1b: SP\r
+ 0x1c: PC\r
+ 0x1d: O\r
+ 0x1e: [next word]\r
+ 0x1f: next word (literal)\r
+ 0x20-0x3f: literal value 0x00-0x1f (literal)\r
+ \r
+* "next word" really means "[PC++]". These increase the word length of the instruction by 1. \r
+* If any instruction tries to assign a literal value, the assignment fails silently. Other than that, the instruction behaves as normal.\r
+* All values that read a word (0x10-0x17, 0x1e, and 0x1f) take 1 cycle to look up. The rest take 0 cycles.\r
+* By using 0x18, 0x19, 0x1a as POP, PEEK and PUSH, there's a reverse stack starting at memory location 0xffff. Example: "SET PUSH, 10", "SET X, POP"\r
+\r
+\r
+\r
+Basic opcodes: (4 bits)\r
+ 0x0: non-basic instruction - see below\r
+ 0x1: SET a, b - sets a to b\r
+ 0x2: ADD a, b - sets a to a+b, sets O to 0x0001 if there's an overflow, 0x0 otherwise\r
+ 0x3: SUB a, b - sets a to a-b, sets O to 0xffff if there's an underflow, 0x0 otherwise\r
+ 0x4: MUL a, b - sets a to a*b, sets O to ((a*b)>>16)&0xffff\r
+ 0x5: DIV a, b - sets a to a/b, sets O to ((a<<16)/b)&0xffff. if b==0, sets a and O to 0 instead.\r
+ 0x6: MOD a, b - sets a to a%b. if b==0, sets a to 0 instead.\r
+ 0x7: SHL a, b - sets a to a<<b, sets O to ((a<<b)>>16)&0xffff\r
+ 0x8: SHR a, b - sets a to a>>b, sets O to ((a<<16)>>b)&0xffff\r
+ 0x9: AND a, b - sets a to a&b\r
+ 0xa: BOR a, b - sets a to a|b\r
+ 0xb: XOR a, b - sets a to a^b\r
+ 0xc: IFE a, b - performs next instruction only if a==b\r
+ 0xd: IFN a, b - performs next instruction only if a!=b\r
+ 0xe: IFG a, b - performs next instruction only if a>b\r
+ 0xf: IFB a, b - performs next instruction only if (a&b)!=0\r
+ \r
+* SET, AND, BOR and XOR take 1 cycle, plus the cost of a and b\r
+* ADD, SUB, MUL, SHR, and SHL take 2 cycles, plus the cost of a and b\r
+* DIV and MOD take 3 cycles, plus the cost of a and b\r
+* IFE, IFN, IFG, IFB take 2 cycles, plus the cost of a and b, plus 1 if the test fails\r
+ \r
+\r
+ \r
+Non-basic opcodes always have their lower four bits unset, have one value and a six bit opcode.\r
+In binary, they have the format: aaaaaaoooooo0000\r
+The value (a) is in the same six bit format as defined earlier.\r
+\r
+Non-basic opcodes: (6 bits)\r
+ 0x00: reserved for future expansion\r
+ 0x01: JSR a - pushes the address of the next instruction to the stack, then sets PC to a\r
+ 0x02-0x3f: reserved\r
+ \r
+* JSR takes 2 cycles, plus the cost of a.\r
+\r
+\r
+\r
+FAQ:\r
+\r
+Q: Why is there no JMP or RET?\r
+A: They're not needed! "SET PC, <target>" is a one-instruction JMP.\r
+ For small relative jumps in a single word, you can even do "ADD PC, <dist>" or "SUB PC, <dist>".\r
+ For RET, simply do "SET PC, POP"\r
+ \r
+Q: How does the overflow (O) work?\r
+A: O is set by certain instructions (see above), but never automatically read. You can use its value in instructions, however.\r
+ For example, to do a 32 bit add of 0x12345678 and 0xaabbccdd, do this:\r
+ SET [0x1000], 0x5678 ; low word\r
+ SET [0x1001], 0x1234 ; high word\r
+ ADD [0x1000], 0xccdd ; add low words, sets O to either 0 or 1 (in this case 1)\r
+ ADD [0x1001], O ; add O to the high word\r
+ ADD [0x1001], 0xaabb ; add high words, sets O again (to 0, as 0xaabb+0x1235 is lower than 0x10000)\r
+\r
+Q: How do I do 32 or 64 bit division using O?\r
+A: This is left as an exercise for the reader.\r
+ \r
+Q: How about a quick example?\r
+A: Sure! Here's some sample assembler, and a memory dump of the compiled code:\r
+\r
+ ; Try some basic stuff\r
+ SET A, 0x30 ; 7c01 0030\r
+ SET [0x1000], 0x20 ; 7de1 1000 0020\r
+ SUB A, [0x1000] ; 7803 1000\r
+ IFN A, 0x10 ; c00d \r
+ SET PC, crash ; 7dc1 001a [*]\r
+ \r
+ ; Do a loopy thing\r
+ SET I, 10 ; a861\r
+ SET A, 0x2000 ; 7c01 2000\r
+ :loop SET [0x2000+I], [A] ; 2161 2000\r
+ SUB I, 1 ; 8463\r
+ IFN I, 0 ; 806d\r
+ SET PC, loop ; 7dc1 000d [*]\r
+ \r
+ ; Call a subroutine\r
+ SET X, 0x4 ; 9031\r
+ JSR testsub ; 7c10 0018 [*]\r
+ SET PC, crash ; 7dc1 001a [*]\r
+ \r
+ :testsub SHL X, 4 ; 9037\r
+ SET PC, POP ; 61c1\r
+ \r
+ ; Hang forever. X should now be 0x40 if everything went right.\r
+ :crash SET PC, crash ; 7dc1 001a [*]\r
+ \r
+ ; [*]: Note that these can be one word shorter and one cycle faster by using the short form (0x00-0x1f) of literals,\r
+ ; but my assembler doesn't support short form labels yet. \r
+\r
+ Full memory dump:\r
+ \r
+ 0000: 7c01 0030 7de1 1000 0020 7803 1000 c00d\r
+ 0008: 7dc1 001a a861 7c01 2000 2161 2000 8463\r
+ 0010: 806d 7dc1 000d 9031 7c10 0018 7dc1 001a\r
+ 0018: 9037 61c1 7dc1 001a 0000 0000 0000 0000\r
+
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